Problem: Alice has a cute cat. To keep her cat fit, Alice wants to design an exercising walk for her cat! 

Initially, Alice's cat is located in a cell (x,y) of an infinite grid. According to Alice's theory, cat needs to move: 

  * exactly a steps left: from (u,v) to (u-1,v); 
  * exactly b steps right: from (u,v) to (u+1,v); 
  * exactly c steps down: from (u,v) to (u,v-1); 
  * exactly d steps up: from (u,v) to (u,v+1). 



Note that the moves can be performed in an arbitrary order. For example, if the cat has to move 1 step left, 3 steps right and 2 steps down, then the walk right, down, left, right, right, down is valid.

Alice, however, is worrying that her cat might get lost if it moves far away from her. So she hopes that her cat is always in the area [x_1,x_2]× [y_1,y_2], i.e. for every cat's position (u,v) of a walk x_1 ≤ u ≤ x_2 and y_1 ≤ v ≤ y_2 holds.

Also, note that the cat can visit the same cell multiple times.

Can you help Alice find out if there exists a walk satisfying her wishes?

Formally, the walk should contain exactly a+b+c+d unit moves (a to the left, b to the right, c to the down, d to the up). Alice can do the moves in any order. Her current position (u, v) should always satisfy the constraints: x_1 ≤ u ≤ x_2, y_1 ≤ v ≤ y_2. The staring point is (x, y).

You are required to answer t test cases independently.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3) — the number of testcases. 

The first line of each test case contains four integers a, b, c, d (0 ≤ a,b,c,d ≤ 10^8, a+b+c+d ≥ 1).

The second line of the test case contains six integers x, y, x_1, y_1, x_2, y_2 (-10^8 ≤ x_1≤ x ≤ x_2 ≤ 10^8, -10^8 ≤ y_1 ≤ y ≤ y_2 ≤ 10^8).

Output

For each test case, output "YES" in a separate line, if there exists a walk satisfying her wishes. Otherwise, output "NO" in a separate line. 

You can print each letter in any case (upper or lower).

Example

Input


6
3 2 2 2
0 0 -2 -2 2 2
3 1 4 1
0 0 -1 -1 1 1
1 1 1 1
1 1 1 1 1 1
0 0 0 1
0 0 0 0 0 1
5 1 1 1
0 0 -100 -100 0 100
1 1 5 1
0 0 -100 -100 100 0


Output


Yes
No
No
Yes
Yes
Yes

Note

In the first test case, one valid exercising walk is $$$(0,0)→ (-1,0) → (-2,0)→ (-2,1) → (-2,2)→ (-1,2)→(0,2)→ (0,1)→ (0,0) → (-1,0)$$$

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class _practise {
	
	static class FastReader
	{ 
		
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				try
				{ 
					st = new StringTokenizer(br.readLine()); 
				} 
				catch (IOException  e) 
				{ 
					e.printStackTrace(); 
				} 
			} 
			return st.nextToken(); 
		} 

		int nextInt() 
		{ 
			return Integer.parseInt(next()); 
		} 

		long nextLong() 
		{ 
			return Long.parseLong(next()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		 int[] ia(int n)
		{
			int a[]=new int[n];
			for(int i=0;i<n;i++)a[i]=nextInt();
			return a;
		}
		 int[][] ia(int n , int m)
			{
				int a[][]=new int[n][m];
				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
				return a;
			}
		 long[][] la(int n , int m)
			{
				long a[][]=new long[n][m];
				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
				return a;
			}
		 char[][] ca(int n , int m)
			{
				char a[][]=new char[n][m];
				for(int i=0;i<n;i++)
					{
					String x =next();
					for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
					}
				return a;
			}

		long[] la(int n)
		{
			long a[]=new long[n];
			for(int i=0;i<n;i++)a[i]=nextLong();
			return a;
		}

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 
	}
	static void sort(long[] a) 
	{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {long oi=r.nextInt(n), temp=a[i];a[i]=a[(int)oi];a[(int)oi]=temp;}Arrays.sort(a);}
    static void sort(int[] a) 
    {int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {int oi=r.nextInt(n), temp=a[i];a[i]=a[oi];a[oi]=temp;}Arrays.sort(a);}
	public static long sum(long a[])
	{long sum=0; for(long i : a) sum+=i; return(sum);}
	public static long count(long a[] , long x)
	{long c=0; for(long i : a) if(i==x) c++; return(c);}
	public static int sum(int a[])
	{ int sum=0; for(int i : a) sum+=i; return(sum);}
	public static int count(int a[] ,int x)
	{int c=0; for(int i : a) if(i==x) c++; return(c);}
	public static int count(String s ,char ch)
	{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
	public static boolean prime(int n)
	{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
	public static boolean prime(long n)
	{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
	 public static int gcd(int n1, int n2)
	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
	 public static long gcd(long n1, long n2)
	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
	 public static int[] freq(int a[], int n)
	  { int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
	 public static int[] pos(int a[], int n)
	  { int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
	 
	
	 public static long mindig(long n)
	 {
		 long ans=9; while(n>0) { if(n%10<ans) ans=n%10; n/=10; }
		 return ans;
	 }
	 public static long maxdig(long n)
	 {
		 long ans=0; while(n>0) { if(n%10>ans) ans=n%10; n/=10; }
		 return ans;
	 }
	  
	public static boolean palin(String s)
	{
		StringBuilder sb = new StringBuilder();
		sb.append(s);
		String str=String.valueOf(sb.reverse());
		if(s.equals(str))
			return true;
		else return false;
	}
	
	public static void main(String args[])
	{

		FastReader in=new FastReader();
		PrintWriter so = new PrintWriter(new BufferedWriter(new  OutputStreamWriter(System.out)));
		_practise ob = new _practise();
		
	    int T = in.nextInt();
		//int T = 1;
	   
	  
	     while(T-->0)
		 {
	    	 
	    	 
	        long a[] = in.la(4);
	        long p[] = in.la(6);
	      
	        long u = p[5]-p[1];
	        long d = p[1]-p[3];
	        long r = p[4]-p[0];
	        long l = p[0]-p[2];
	     // so.println(x+" "+y);
	      // so.println(u+" "+d+" "+r+" "+l);
	     
	        if((l+r==0 && a[0]+a[1]!=0)||(u+d==0 && a[2]+a[3]!=0))so.println("NO");
	        else if((a[2]>a[3]&&a[2]-a[3]>d) || (a[1]>a[0]&&a[1]-a[0]>r)) so.println("NO");
	        else if((a[3]>a[2] && a[3]-a[2]>u) || (a[0]>a[1]&&a[0]-a[1]>l)) so.println("NO");
	        else so.println("YES");
	        
		 }
		
		so.flush();

		/*String s = in.next();
		 * Arrays.stream(f).min().getAsInt()
		 * BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap 
		 *                                       initial value banan chahte
			int a[] = new int[n];
			 Stack<Integer> stack = new Stack<Integer>();
			 	Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
			PriorityQueue<Long> pq = new PriorityQueue<Long>();
			ArrayList<Integer> al = new ArrayList<Integer>(); 
			StringBuilder sb = new StringBuilder();
			HashSet<Integer> st = new LinkedHashSet<Integer>();
			Set<Integer> s = new HashSet<Integer>();
			Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
			 for(Map.Entry<Integer, Integer> i :hm.entrySet())
			HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
			so.println("HELLO");
			Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
			 Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));*/
	}
}